Xét Δ`ABC` ta có:
$\widehat{A}$`+`$\widehat{B}$`+`$\widehat{C}$`=180^0`(Tổng 3 góc trong 1 tam giác)
`=>`$\widehat{B}$`+`$\widehat{C}$ `= 180^0 -` $\widehat{A}$
Vì `BI` là tia phân giác của $\widehat{B}$
`=>` $\widehat{CBI}$`=1/2` $\widehat{B}$
Vì `CI` là tia phân giác của $\widehat{C}$
`=>` $\widehat{CBI}$`=1/2` $\widehat{B}$
Xét Δ`BIC` ta có:
$\widehat{CBI}$`+`$\widehat{ICB}$`+`$\widehat{BIC}$`=180^0` (Tổng 3 góc trong 1 tam giác)
`=>`$\widehat{CBI}$`+`$\widehat{ICB}$`=180^0``-`$\widehat{BIC}$
Hay `1/2` $\widehat{B}$`+``1/2`$\widehat{C}$`=180^0``-`$\widehat{BIC}$
`=>` `1/2`($\widehat{B}$`+`$\widehat{C}$)`=180^0``-`$\widehat{BIC}$
`=>` `1/2`(`180^0` `-` $\widehat{A}$)`=180^0``-`$\widehat{BIC}$
`=>` `90^0` `-` $\frac{\widehat{A}}{2}$ `=180^0``-`$\widehat{BIC}$
`=> 180^0 -` (`90^0` `-` $\frac{\widehat{A}}{2}$)`=`$\widehat{BIC}$
`=> 180^0 - 90^0 +` $\frac{\widehat{A}}{2}$ `=` $\widehat{BIC}$
`=> 90^0 +` $\frac{\widehat{A}}{2}$ `=` $\widehat{BIC}$
`=>`$\widehat{BIC}$`=90^0` `+` $\frac{\widehat{A}}{2}$
Vậy $\widehat{BIC}$`=90^0` `+` $\frac{\widehat{A}}{2}$
